Python：将2d二进制值数组打包为UINT64数组的最快方法

Question

我有一个2D UINT8 numpy数组，大小为（（149797，64）]。每个元素都是0或1。我想将每行中的这些二进制值打包到UINT64值中，以便得到一个形状为149797的UINT64数组。我使用numpy bitpack函数尝试了以下代码。

test = np.random.randint(0, 2, (149797, 64),dtype=np.uint8)
col_pack=np.packbits(test.reshape(-1, 8, 8)[:, ::-1]).view(np.uint64)

packbits函数大约需要10 ms才能执行。这个数组本身的简单reshaping似乎要花费7 ms左右。我还尝试了通过移位操作在2d numpy数组上进行迭代以达到相同的结果。但速度没有改善。

最后，我也想使用numba进行编译。

@njit
def shifting(bitlist):
    x=np.zeros(149797,dtype=np.uint64)  #54
    rows,cols=bitlist.shape
    for i in range(0,rows):             #56
      out=0
      for bit in range(0,cols):
         out = (out << 1) | bitlist[i][bit] # If i comment out bitlist, time=190 microsec
      x[i]=np.uint64(out)  # Reduces time to microseconds if line is commented in njit
    return x

使用njit大约需要6毫秒。

这里是并行的njit版本

@njit(parallel=True)
def shifting(bitlist): 
    rows,cols=149797,64
    out=0
    z=np.zeros(rows,dtype=np.uint64)
    for i in prange(rows):
      for bit in range(cols):
         z[i] = (z[i] << 1) | bitlist[i,bit] # Time becomes 100 micro if i use 'out' instead of 'z[i] array'

    return z

与5.6ms执行时间相比要好一些。当前，使用swapbytes（Paul's）方法的python解决方案似乎是最好的方法，即1.74 ms。

我们如何进一步加快转换速度？是否可以使用任何矢量化（或并行化），位数组等来实现加速？

Ref：numpy packbits pack to uint16 array

Answer 1

您可以通过使用byteswap而不是重新塑形来获得可观的加速：

test = np.random.randint(0, 2, (149797, 64),dtype=np.uint8)

np.packbits(test.reshape(-1, 8, 8)[:, ::-1]).view(np.uint64)
# array([ 1079982015491401631,   246233595099746297, 16216705265283876830,
#        ...,  1943876987915462704, 14189483758685514703,
       12753669247696755125], dtype=uint64)
np.packbits(test).view(np.uint64).byteswap()
# array([ 1079982015491401631,   246233595099746297, 16216705265283876830,
#        ...,  1943876987915462704, 14189483758685514703,
       12753669247696755125], dtype=uint64)

timeit(lambda:np.packbits(test.reshape(-1, 8, 8)[:, ::-1]).view(np.uint64),number=100)
# 1.1054180909413844

timeit(lambda:np.packbits(test).view(np.uint64).byteswap(),number=100)
# 0.18370431219227612

Answer 2

有点Numba解决方案（版本0.46 / Windows）。

代码

import numpy as np
import numba as nb

#with memory allocation
@nb.njit(parallel=True)
def shifting(bitlist):
    assert bitlist.shape[1]==64
    x=np.empty(bitlist.shape[0],dtype=np.uint64)

    for i in nb.prange(bitlist.shape[0]):
        out=np.uint64(0)
        for bit in range(bitlist.shape[1]):
            out = (out << 1) | bitlist[i,bit] 
        x[i]=out
    return x

#without memory allocation
@nb.njit(parallel=True)
def shifting_2(bitlist,x):
    assert bitlist.shape[1]==64

    for i in nb.prange(bitlist.shape[0]):
        out=np.uint64(0)
        for bit in range(bitlist.shape[1]):
            out = (out << 1) | bitlist[i,bit] 
        x[i]=out
    return x

Timings

test = np.random.randint(0, 2, (149797, 64),dtype=np.uint8)

#If you call this function multiple times, only allocating memory 
#once may be enough
x=np.empty(test.shape[0],dtype=np.uint64)

#Warmup first call takes significantly longer
res=shifting(test)
res=shifting_2(test,x)

%timeit res=shifting(test)
#976 µs ± 41.6 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
%timeit res=shifting_2(test,x)
#764 µs ± 63 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
%timeit np.packbits(test).view(np.uint64).byteswap()
#8.07 ms ± 52.5 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
%timeit np.packbits(test.reshape(-1, 8, 8)[:, ::-1]).view(np.uint64)
#17.9 ms ± 91 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)